Circuit analysis - Finding the currents

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Homework Help Overview

The discussion revolves around analyzing a circuit to find three unknown currents using Kirchhoff's laws. Participants are examining the setup and equations derived from the circuit, questioning the assumptions made regarding current directions and voltage drops.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of Kirchhoff's laws and the derivation of equations based on potential walks around the circuit. Questions arise about the sign conventions used in the equations and the reasoning behind current directions, particularly for I_2.

Discussion Status

The discussion is active, with participants exploring different interpretations of the circuit's behavior. Some guidance on sign conventions has been provided, but there is still uncertainty regarding the assumptions about current directions and voltage levels.

Contextual Notes

There is a mention of potential ambiguity in the problem if voltage values are not explicitly provided, which could affect the determination of current directions.

hadoque
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Homework Statement



Find the three currents in the image.

Homework Equations


Khirkoffs laws
Sum of currents equal zero.
The potentials around a closed loop add up to zero.
The equations given in the solution are
[tex]-2I_1 + 12I_2+6=0[/tex]
[tex]-12I_2+8 = 0[/tex]
[tex]I_1+I_2+I_3=0[/tex]

The Attempt at a Solution



When I tried to solve this I had the first equation to be
[tex]-2I_1-12I_2+6=0[/tex], by doing a potential walk round the left subcircuit (?).
Why should the second term be positive?
 

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hadoque said:

Homework Statement



Find the three currents in the image.

Homework Equations


Khirkoffs laws
Sum of currents equal zero.
The potentials around a closed loop add up to zero.
The equations given in the solution are
[tex]-2I_1 + 12I_2+6=0[/tex]
[tex]-12I_2+8 = 0[/tex]
[tex]I_1+I_2+I_3=0[/tex]

The Attempt at a Solution



When I tried to solve this I had the first equation to be
[tex]-2I_1-12I_2+6=0[/tex], by doing a potential walk round the left subcircuit (?).
Why should the second term be positive?

By convention, when you have the current going down through a component, the top of the component will be + and the bottom will be -. Since you are going around the loop on the left against the direction labelled for I1, it is a negative voltage term. Since you are continuing around that clockwise loop on the left and go through the middle resistor in the same direction as the current arrow shown, it is a + voltage drop.
 
Ok, but how do I know the current is going down in I_2, besides the arrow? If the problem didn't give the currents directions, I should still be able to solve it! Is it that the right battery has a higher voltage? If so, what if the problem didn't give the battery voltage in numerics, but instead gave, say, the voltage v_1 and v_2. Then I wouldn't know which one has the higher voltage...
 
Ok, sorry. i just realized I was thinking wrong about the circuit, there is only one option for the direction of I_2.

Thanks
 

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